miljanwpd2

2023-03-23

What is the projection of $<6,5,3>$ onto $<2,-1,8>$?

Saottah87e

Beginner2023-03-24Added 6 answers

Given $\overrightarrow{a}=<6,5,3>$ and $\overrightarrow{b}=<2,-1,8>$, we can find $pro{j}_{\overrightarrow{b}}\overrightarrow{a}$, the vector projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ using the following formula:

$pro{j}_{\overrightarrow{b}}\overrightarrow{a}=\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right)\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$

That is, the dot product of the two vectors divided by the magnitude of $\overrightarrow{b}$, multiplied by $\overrightarrow{b}$ divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide $\overrightarrow{b}$ by its magnitude in order to obtain a unit vector (vector with magnitude of $1$). You may have noticed that the first quantity is a scalar because we know that the dot product of two vectors always yields a scalar value.

Therefore, the scalar projection of $a$ onto $b$ is $com{p}_{\overrightarrow{b}}\overrightarrow{a}=\frac{a\cdot b}{\left|b\right|}$, also written $\left|pro{j}_{\overrightarrow{b}}\overrightarrow{a}\right|$

Taking the dot product of the two vectors will be our first step.

$\overrightarrow{a}\cdot \overrightarrow{b}=<6,5,3>\cdot <2,-1,8>$

$\Rightarrow (6\cdot 2)+(5\cdot -1)+(3\cdot 8)$

$\Rightarrow 12-5+24=31$

Then, by taking the square root of the sum of the squares of each component, we can determine the size of $\overrightarrow{b}$

$\left|\overrightarrow{b}\right|=\sqrt{{\left({b}_{x}\right)}^{2}+{\left({b}_{y}\right)}^{2}+{\left({b}_{z}\right)}^{2}}$

$\left|\overrightarrow{b}\right|=\sqrt{{\left(2\right)}^{2}+{(-1)}^{2}+{\left(8\right)}^{2}}$

$\Rightarrow \sqrt{4+1+64}=\sqrt{69}$

And now we have everything we need to find the vector projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$

$pro{j}_{\overrightarrow{b}}\overrightarrow{a}=\frac{31}{\sqrt{69}}\cdot \frac{<2,-1,8>}{\sqrt{69}}$

$\Rightarrow \frac{31<2,-1,8>}{69}$

$=\frac{31}{69}<2,-1,8>$

The coefficient can be distributed to each element of the vector and written as follows:

$\Rightarrow <\frac{62}{69},-\frac{31}{69},\frac{248}{69}>$

The scalar projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ is just the first half of the formula, where $com{p}_{\overrightarrow{b}}\overrightarrow{a}=\frac{a\cdot b}{\left|b\right|}$. Therefore, the scalar projection is $\frac{31}{\sqrt{69}}$, which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{31\sqrt{69}}{69}$

$pro{j}_{\overrightarrow{b}}\overrightarrow{a}=\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right)\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$

That is, the dot product of the two vectors divided by the magnitude of $\overrightarrow{b}$, multiplied by $\overrightarrow{b}$ divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide $\overrightarrow{b}$ by its magnitude in order to obtain a unit vector (vector with magnitude of $1$). You may have noticed that the first quantity is a scalar because we know that the dot product of two vectors always yields a scalar value.

Therefore, the scalar projection of $a$ onto $b$ is $com{p}_{\overrightarrow{b}}\overrightarrow{a}=\frac{a\cdot b}{\left|b\right|}$, also written $\left|pro{j}_{\overrightarrow{b}}\overrightarrow{a}\right|$

Taking the dot product of the two vectors will be our first step.

$\overrightarrow{a}\cdot \overrightarrow{b}=<6,5,3>\cdot <2,-1,8>$

$\Rightarrow (6\cdot 2)+(5\cdot -1)+(3\cdot 8)$

$\Rightarrow 12-5+24=31$

Then, by taking the square root of the sum of the squares of each component, we can determine the size of $\overrightarrow{b}$

$\left|\overrightarrow{b}\right|=\sqrt{{\left({b}_{x}\right)}^{2}+{\left({b}_{y}\right)}^{2}+{\left({b}_{z}\right)}^{2}}$

$\left|\overrightarrow{b}\right|=\sqrt{{\left(2\right)}^{2}+{(-1)}^{2}+{\left(8\right)}^{2}}$

$\Rightarrow \sqrt{4+1+64}=\sqrt{69}$

And now we have everything we need to find the vector projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$

$pro{j}_{\overrightarrow{b}}\overrightarrow{a}=\frac{31}{\sqrt{69}}\cdot \frac{<2,-1,8>}{\sqrt{69}}$

$\Rightarrow \frac{31<2,-1,8>}{69}$

$=\frac{31}{69}<2,-1,8>$

The coefficient can be distributed to each element of the vector and written as follows:

$\Rightarrow <\frac{62}{69},-\frac{31}{69},\frac{248}{69}>$

The scalar projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ is just the first half of the formula, where $com{p}_{\overrightarrow{b}}\overrightarrow{a}=\frac{a\cdot b}{\left|b\right|}$. Therefore, the scalar projection is $\frac{31}{\sqrt{69}}$, which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{31\sqrt{69}}{69}$

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